leetcode_question_101 Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).For example, this binary tree is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following is not:
1
/ \
2 2
\ \
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
Recursive:
[cpp]
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool symmetric(TreeNode * left, TreeNode* right)
{
if(left== NULL && right == NULL)return true;
if(left && right && left->val == right->val)
return symmetric(left->right,right->left) && symmetric(left->left, right->right);
else return false;
}
bool isSymmetric(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(root==NULL)return true;
return symmetric(root->left,root->right);
}
};
中序遍历：
[cpp]
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void midvisit(TreeNode* root, vector<int>&path)
{
if(root)
{
midvisit(root->left,path);
path.push_back(root->val);
midvisit(root->right,path);
}
}
bool isSymmetric(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(root==NULL)return true;
vector<int> path;
midvisit(root,path);
int length = path.size();
bool flag = true;
for(int i = 0, j = length-1; i < j; ++i, --j)
{
if(path[i] != path[j]) {flag = false;break;}
}
return flag;
}
};
Iterative:
[cpp]
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isSymmetric(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(root==NULL)return true;
queue<TreeNode*> lt,rt;
if(root->left) lt.push(root->left);
if(root->right) rt.push(root->right);
TreeNode* l;
TreeNode* r;
while(!lt.empty() && !rt.empty())
{
l = lt.front();lt.pop();
r = rt.front();rt.pop();
if(l == NULL && r == NULL) continue;
if(l == NULL || r == NULL) return false;
if(l->val != r->val) return false;
lt.push(l->left);
lt.push(l->right);
rt.push(r->right);
rt.push(r->left);
}
if(lt.empty() && rt.empty())
return true;
else
return false;
}
};
补充：软件开发 , C++ ,
